(Mon) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point,

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Presentation transcript:

(Mon) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point, how much work was done to get the car up here? (5 min/5 pts)

100 m W = F g d W = ma g d W = (1000 kg)(9.81 m/s²)(100 m) 9.81 x 10 5 J

(Tue) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point and is rolling with a speed of 10 m/s, what is its mechanical energy at the top of the hill? (5 min/5pts)

100 m ME = KE + ΣPE ME = ½mv² + mgh + ½kx² ME = ½(1000 kg)(10 m/s)² + (1000 kg)(9.81 m/s²)(100 m) + 0 ME = 50 kJ kJ ME = J = 1031 kJ ME = MJ 10 m/s

(Wed) A roller coaster car loaded with people has a mass of 1000 kg. If the car at the top of the first hill has an initial ME of MJ, and it drops 130 m on the first hill (30 m below the starting point) what is its speed when it reaches the bottom of the first drop (neglect friction)? (5 min/5 pts)

130 m MJ MEi = MEf MJ = ½mv² + mgh + ½kx² MJ = ½(1000 kg)v² + (1000 kg)(9.81 m/s²)(-30 m) MJ = 500v² + ( J) (1.031 MJ MJ) / (500) = v² v² = m²/s² v = m/s 30 m

(Thu) A roller coaster car loaded with people has a mass of 1000 kg. Your car with a total ME of MJ at the top of the first hill climbs from the first drop to the top of the second hill. If your speed at the top of the second hill is m/s, how high above the starting point is the second hill (neglect friction)? (5 min/5 pts)

1.031 MJ ? m MEi = MEf MJ = ½mv² + mgh + ½kx² (Thu) A roller coaster car loaded with people has a mass of 1000 kg. Your car with a total ME of MJ at the top of the first hill climbs from the first drop to the top of the second hill. If your speed at the top of the second hill is m/s, how high above the starting point is the second hill (neglect friction)? (5 min/5 pts) MJ = ½(1000 kg)(32.88 m/s)² + (1000 kg)(9.81 m/s²)(h) MJ = J + (9810 kgm/s²)(h) (1.031 MJ) – ( J) / (9810 kgm/s²) = h h = 50 m

(Fri) Your cell phone slips out of your hand at the top of the second hill on the roller coaster. If you are traveling horizontally at a speed of m/s and are 50 m above the ground when you lose it, how far from a point directly under the second hill should you start looking for your cell phone? (10 min/8 pts)

(Fri) Your cell phone slips out of your hand at the top of the second hill on the roller coaster. If you are traveling horizontally at a speed of m/s and are 50 m above the ground when you lose it, how far from a point directly under the second hill should you start looking for your cell phone? m/s 50 m x Diry Dir ΔxΔx vivi vfvf a t How long does it take to fall? (y direction) ??? Δx = v i t + ½at² -50m = 0 + ½(-9.81m/s²)t² (50/4.905)s² = t² t = 3.19s 6.71s How far does it go horizontally? (x direction) ??? s Δx = vtΔx = (32.88 m/s)(3.19 s) Δx = m (8 pts) If the cell phone has a mass of 100 g, how fast is the cell phone going when it hits the ground? (5 min) ME i = ME f ½mv i ² + mgh i = ½mv f ² + mgh f ½(0.1)(32.88)² + (0.1)(9.81)(50) = ½(0.1)v f ² + 0 v f =45.41 m/s (5 pts) ( )(2)/(0.1)=v f ²

End of Week Procedures: 1.Add up all the points you got this week 2.Put the total number of points at the top of your page (out of 28 points) 3.List any dates you were absent (and why) 4.Make sure your name and period is on the top of the paper 5.Turn in your papers